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#include <bits/stdc++.h>
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using namespace std;
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const int INF = 0x3f3f3f3f;
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// dfs只能求出来是否连通,第一次搜索到时并不能保证是最短距离
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// bfs也可以做,可以保证第一次到达时是最短距离
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// dfs好处是代码短,按时间排名,那么先AC的同学排名靠前
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// 用标记数组进行标记,每个位置只使用一次,性能N*N
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const int N = 110;
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int n;
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char g[N][N]; // 地图
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int xa, ya, xb, yb;
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int dx[] = {-1, 0, 1, 0};
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int dy[] = {0, 1, 0, -1};
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bool st[N][N]; // 是否走过
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bool flag;
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void dfs(int x, int y) {
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if (x == xb && y == yb) {
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flag = true;
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return;
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}
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for (int i = 0; i < 4; i++) {
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int tx = x + dx[i], ty = y + dy[i];
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if (tx < 0 || tx == n || ty < 0 || ty == n) continue;
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if (st[tx][ty]) continue;
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if (g[tx][ty] == '#') continue;
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st[tx][ty] = true;
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dfs(tx, ty);
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}
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}
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int main() {
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int T;
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cin >> T;
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while (T--) {
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cin >> n;
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for (int i = 0; i < n; i++) cin >> g[i];
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cin >> xa >> ya >> xb >> yb;
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// 多组测试数组,每次初始化0
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memset(st, 0, sizeof st);
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flag = false;
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// 这小坑坑挺多啊
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if (g[xa][ya] == '#' || g[xb][yb] == '#') {
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puts("NO");
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continue;
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}
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st[xa][ya] = true;
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dfs(xa, ya);
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if (flag)
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puts("YES");
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else
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puts("NO");
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}
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return 0;
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}
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